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Proof step 13 from proof byte2bits-val (previous step)

# int2bits(b2i(by)) ≠ 0, 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ bits2uint(mklist(true, n)) * 2 ^ n→i(# int2bits(b2i(by))) + bits2uint(int2bits(b2i(by))) = b2i(by) + 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];